Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 110;
int n, m;
int G[MAX][MAX];
vector<int> v_row;
vector<int> v_col;
void find_row_max_min(int r, int& v_max, int& v_min) {
  for (int c = 0; c < m; ++c) {
    v_max = max(v_max, G[r][c]);
    v_min = min(v_min, G[r][c]);
  }
}
void find_col_max_min(int c, int& v_max, int& v_min) {
  for (int r = 0; r < n; ++r) {
    v_max = max(v_max, G[r][c]);
    v_min = min(v_min, G[r][c]);
  }
}
void solve_1() {
  bool is_ok = 1;
  for (int r = 0; r < n; ++r) {
    int v_max = 0, v_min = 501;
    find_row_max_min(r, v_max, v_min);
    for (int i = 0; i < v_min; ++i) {
      v_row.push_back(r + 1);
    }
    for (int c = 0; c < m; ++c) {
      G[r][c] -= v_min;
    }
    if (v_max != v_min) {
      is_ok = 0;
    }
  }
  if (is_ok) {
    cout << v_row.size() << endl;
    for (int i = 0; i < v_row.size(); ++i) {
      cout << "row " << v_row[i] << endl;
    }
    return;
  }
  is_ok = 1;
  for (int c = 0; c < m; ++c) {
    int v_max = 0, v_min = 501;
    find_col_max_min(c, v_max, v_min);
    for (int i = 0; i < v_min; ++i) {
      v_col.push_back(c + 1);
    }
    for (int r = 0; r < n; ++r) {
      G[r][c] -= v_min;
    }
    if (v_max != v_min) {
      is_ok = 0;
    }
  }
  if (is_ok) {
    cout << v_row.size() + v_col.size() << endl;
    for (int i = 0; i < v_row.size(); ++i) {
      cout << "row " << v_row[i] << endl;
    }
    for (int i = 0; i < v_col.size(); ++i) {
      cout << "col " << v_col[i] << endl;
    }
    return;
  }
  cout << "-1" << endl;
  return;
}
void solve_2() {
  bool is_ok = 1;
  for (int c = 0; c < m; ++c) {
    int v_max = 0, v_min = 501;
    find_col_max_min(c, v_max, v_min);
    for (int i = 0; i < v_min; ++i) {
      v_col.push_back(c + 1);
    }
    for (int r = 0; r < n; ++r) {
      G[r][c] -= v_min;
    }
    if (v_max != v_min) {
      is_ok = 0;
    }
  }
  if (is_ok) {
    cout << v_col.size() << endl;
    for (int i = 0; i < v_col.size(); ++i) {
      cout << "col " << v_col[i] << endl;
    }
    return;
  }
  is_ok = 1;
  for (int r = 0; r < n; ++r) {
    int v_max = 0, v_min = 501;
    find_row_max_min(r, v_max, v_min);
    for (int i = 0; i < v_min; ++i) {
      v_row.push_back(r + 1);
    }
    for (int c = 0; c < m; ++c) {
      G[r][c] -= v_min;
    }
    if (v_max != v_min) {
      is_ok = 0;
    }
  }
  if (is_ok) {
    cout << v_row.size() + v_col.size() << endl;
    for (int i = 0; i < v_row.size(); ++i) {
      cout << "row " << v_row[i] << endl;
    }
    for (int i = 0; i < v_col.size(); ++i) {
      cout << "col " << v_col[i] << endl;
    }
    return;
  }
  cout << "-1" << endl;
  return;
}
int main() {
  cin >> n >> m;
  for (int r = 0; r < n; ++r)
    for (int c = 0; c < m; ++c) cin >> G[r][c];
  if (n <= m)
    solve_1();
  else
    solve_2();
  return 0;
}